How many moles of `KMnO_(4)` are needed to oxidise a mixture of 1 mole of each `FeSO_(4) & FeC_(2)O_(4)` in acidic medium `:`

Options

A. `4//5`
B. `5//4`
C. `3//4`
D. `5//3`

Explanation

To determine how many moles of KMnO4 are needed to oxidize a mixture of 1 mole of FeSO4 and 1 mole of FeC2O4 in acidic medium, we can follow these steps:

### Step 1: Write the half-reaction for KMnO4
In acidic medium, KMnO4 acts as an oxidizing agent and gets reduced from MnO4^- (where Mn is in +7 oxidation state) to Mn^2+ (where Mn is in +2 oxidation state). The half-reaction can be written as:
\[ \text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} \]

### Step 2: Determine the oxidation reactions for FeSO4 and FeC2O4
- **For FeSO4**: Iron (Fe) is oxidized from +2 to +3 oxidation state:
\[ \text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^- \]
The n-factor for FeSO4 is 1 (1 mole of electrons per mole of Fe).

- **For FeC2O4**: The carbon in oxalic acid (C2O4^2-) is oxidized from +3 to +4 oxidation state:
\[ \text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^- \]
The n-factor for FeC2O4 is 2 (2 moles of electrons per mole of C).

### Step 3: Calculate the total equivalents of the mixture
- **Equivalents of FeSO4**:
\[ \text{Equivalents of FeSO4} = \text{moles} \times \text{n-factor} = 1 \times 1 = 1 \]

- **Equivalents of FeC2O4**:
\[ \text{Equivalents of FeC2O4} = \text{moles} \times \text{n-factor} = 1 \times 2 = 2 \]

### Step 4: Total equivalents in the mixture
\[ \text{Total equivalents in the mixture} = \text{Equivalents of FeSO4} + \text{Equivalents of FeC2O4} = 1 + 2 = 3 \]

### Step 5: Relate the equivalents of KMnO4 to the equivalents of the mixture
The equivalents of KMnO4 will equal the total equivalents of the mixture:
\[ \text{Equivalents of KMnO4} = \text{Total equivalents in the mixture} = 3 \]

### Step 6: Calculate moles of KMnO4 needed
Using the n-factor of KMnO4, which is 5 (as it accepts 5 electrons):
\[ \text{Equivalents of KMnO4} = \text{moles of KMnO4} \times \text{n-factor of KMnO4} \]
Let \( x \) be the moles of KMnO4:
\[ 3 = x \times 5 \]
\[ x = \frac{3}{5} \]

### Conclusion
Thus, the moles of KMnO4 needed to oxidize the mixture of 1 mole of FeSO4 and 1 mole of FeC2O4 is:
\[ \boxed{\frac{3}{5}} \]